Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. For example: Input: "abab" Output: True Explanation: It's the substring "ab" twice. Input: "aba" Output: False Input: "abcabcabcabc" Output: True Explanation: It's the substring "abc" four times. Reference: https://leetcode.com/problems/repeated-substring-pattern/description/ """ def repeat_substring(s): """ :type s: str.
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Given a non-empty string check if it can be constructed by taking | |
a substring of it and appending multiple copies of the substring together. | |
For example: | |
Input: "abab" | |
Output: True | |
Explanation: It's the substring "ab" twice. | |
Input: "aba" | |
Output: False | |
Input: "abcabcabcabc" | |
Output: True | |
Explanation: It's the substring "abc" four times. | |
Reference: https://leetcode.com/problems/repeated-substring-pattern/description/ | |
""" | |
def repeat_substring(s): | |
""" | |
:type s: str.
|
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- Correct answer will be upvoted else Multiple Downvoted. Computer science. You are given a string s, consisting of brackets of two types: '(', ')', '[' and ']'. A string is called a regular bracket sequence (RBS) if it's of one of the following type empty strin '(' + RBS + ')' '[' + RBS + ']' RBS + RBS where plus is a concatenation of two strings In one move you can choose a non-empty subsequence of the string s (not necessarily consecutive) that is an RBS, remove it from the string and concatenate the remaining parts without changing the order What is the maximum number of moves you can perfor Input The first line contains a single integer t (1≤t≤1000) — the number of testcases Each of the next t lines contains a non-empty string, consisting only of characters '(', ')', '[' and ']'. The total length of the strings over all testcases doesn't exceed 2⋅10 Output For each testcase print a single integer — the maximum number of moves you can perform on a given string…Draw a DFA that accepts strings that do not contain two consecutive a's or two consecutive b's. Here, Σ = {a,b}. e.g. aba, bab, ababa, bababa, etc.Check whether a given string is Heterogram or not. Hint: A heterogram is a word, phrase, or sentence in which no letter of the alphabet occurs more than once
- Write in C Language Spilitology Yosef is a peculiar fellow. He introduced the idea to study a string by splitting it into two, and he called it Splitology. Why split a string? We do not know. Didn’t we say that Yosef is a weird one? Yosef is interested in one particular type of string, a palindrome. A palindrome is a string that is the same for both forwards and backwards. Example of palindrome strings are “ada”, “taat”, and “radar”. On the other hand, string such as “taman” is not a palindrome; notice that “taman” becomes “namat” if read backwardsand it’s not the same as “taman”. As the idea of Splitology is still new, Yosef is investigating whether a string can be split into two non-empty strings such that each string is a palindrome. For example, the string “malamini” can be split into “malam” and “ini” while both of them are palindrome. Another example is “ababab”. It can be split into “aba” and “bab”, and both of them are palindrome. Note that “ababab” can also be split into…public static String pancakeScramble(String text) This nifty little problem is taken from the excellent Wolfram Challenges problem site where you can also see examples of what the result should be for various arguments. Given a text string, construct a new string by reversing its first two characters, then reversing the first three characters of that, and so on, until the last round where you reverse your entire current string.This problem is an exercise in Java string manipulation. For some mysterious reason, the Java String type does not come with a reverse method. The canonical way to reverse a Java string str is to first convert it to mutable StringBuilder, reverse its contents, and convert the result back to an immutable string, that is,str = new StringBuilder(str).reverse().toString(); Here's the tester it must pass: @Test public void testPancakeScramble() throws IOException {// Explicit test casesassertEquals("", P2J3.pancakeScramble(""));assertEquals("alu",…public static String pancakeScramble(String text) This nifty little problem is taken from the excellent Wolfram Challenges problem site where you can also see examples of what the result should be for various arguments. Given a text string, construct a new string by reversing its first two characters, then reversing the first three characters of that, and so on, until the last round where you reverse your entire current string. This problem is an exercise in Java string manipulation. For some mysterious reason, the Java String type does not come with a reverse method. The canonical way to reverse a Java string str is to first convert it to mutable StringBuilder, reverse its contents, and convert the result back to an immutable string, that is, str = new StringBuilder(str).reverse().tostring(); A bit convoluted, but does what is needed without fuss or muss. Maybe one day the Java strings will come with the reverse method built in, just like the string data types of all sensible…
- Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1. Example 1: Input: s= "leetcode" Output: 0 Example 2: Input: s= "loveleetcode" Output: 2 Example 3: Input: s= "aabb" Output: -1DescriptionA researcher is analyzing DNA. A DNA can be represented as a string composed of the characters A, G, C, or T.One day, researchers found a strange DNA, which is Smooth Repeated DNA. The DNA is represented by a string that has infinite length. The string has a repeating pattern, i.e. the DNA string 0 is repeated an infinite number of times. For example, if0 = "????", then = "???????????? . . . ".According to researchers, a DNA is said to be special if it contains substrings . Determine whetheris a substring of . Squad FormatA line containing the two strings 0 and . Output FormatA line that determines whether it is a substring of . Issue “YES” ifis a substring of . Output “NO” otherwise. Example Input and Output Input Example Example Output AGCT GC YES AGCT TA YES AGCT GT No AGCT TAGCTAGCT YES AGGACCTA CTAA YES Explanation ExampleIn the first to fourth test case examples, is worth "???????????? . . . ". The part in bold is one of the…if strings are immutable, like in Java, we can assume that they have a static length True False
- Palindromes - “A palindrome” is a string that reads the same from both directions. For example: the word "mom" is a palindrome. Also, the string "Murder for a jar of red rum" is a palindrome. - So, you need to implement a Boolean function that takes as input a string and its return is true (1) in case the string is a palindrome and false (0) otherwise. - There are many ways to detect if a phrase is a palindrome. The method that you will implement in this task is by using two stacks. This works as follows. Push the left half of the characters to one stack (from left to right) and push the second half of the characters (from right to left) to another stack. Pop from both stacks and return false if at any time the two popped characters are different. Otherwise, you return true after comparing all the elements. Phrases of odd length have to be treated by skipping the middle element like the word "mom", your halves are "m" and "m". - Hint: (without using STL)Write a program that reads a string and checks the longest substring starting with "a" and ending with "z" included in it. Note that, the substring cannot contain another a or another z in it. For example: Ifs="bcd bbccdd badefgze tabdz", the longest substring is: 6Ex: Use getline() to get a line of user input into a string. Output the line. Enter text: IDK if I'll go. It's my BFF's birthday. You entered: IDK if I'll go. It's my BFF's birthday. Search the string (using find()) for common abbreviations and print a list of each found abbreviation along with its decoded meaning. Ex: Enter text: IDK if I'll go. It's my BFF's birthday. You entered: IDK if I'll go. It's my BFF's birthday. BFF: best friend forever IDK: I don't know Support these abbreviations: BFF -- best friend forever IDK -- I don't know . JK -- just kidding TMI -- too much information TTYL-talk to you later 1 #include 2 // FIXME include the string Library 3 using namespace std; 4 5 int main() { 6 7 8 9 10 } /* Type your code here. */ return 0; main.cpp